Straightforward vector calc problem

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Studying for finals and I don't recall how to do projection and I don't like the explanation my textbook gives

Problem: Find the distance from the point (1,3,-2) to the plane 2x + y - z = 1

The answer to this problem should be sqrt(6)

Relevant equations: ?

The attempt at a solution:

I know I have to do a projection of using two vectors, and I know one of the vectors is the normal vector to the plane (2i + j + k). I'm not sure how to find the other vector using the point given and which of the vectors is projected onto the other.
 
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Well, what equation / method are you using to find the distance from the point to the plane?
 
draw a picture first...
consider an arbitrary point P on the plane, let Q=(1,3,-2)
now the magnitude of the proj of PQ in the direction of your normal vector IS the distance from Q to plane...
 
ok, thanks. I got it now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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